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[SOLVED] Here's a cute predicate logic problem.
Prove or disprove: \[\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw\]Assume that \(x, y, z, w \in \mathbb{C}\).
 one year ago
 one year ago
[SOLVED] Here's a cute predicate logic problem. Prove or disprove: \[\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw\]Assume that \(x, y, z, w \in \mathbb{C}\).
 one year ago
 one year ago

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experimentXBest ResponseYou've already chosen the best response.0
is xyzw a combination of digit or multiplication of digit??
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
lol .. didn't realize they were complex
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
It doesn't really matter. I just put that specification on there so that know one tried to change the problem so that addition and multiplication had different meanings.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
Just prove/disprove it for \(\mathbb{R}\) and it will become clear why it doesn't matter if they're complex numbers.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
Hint below in white LaTeX\[\color{white}{\text{There is a certain value for y that you should choose to prove it.}}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ \color{white}{\text{LOL i didn't know that!!.}} \]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
Another, probably more helpful hint, below in white LaTeX\[\color{white}{\text{What value of y can you choose so that xyzw is constant?}}\]
 one year ago

nbouscalBest ResponseYou've already chosen the best response.2
Choose y=0, then choose w=(x+z).
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
But ∃y doesn't mean that y is rather a variable such that we have to fit constraint ??
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
\(\exists\, y\) means that we can choose a specific value for y. It's a variable, but one that we have complete control over.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
then it must be false ... i guess so!!
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
The "there exists" only means that at least one possible value of y where this is true. One value that we know works is \(y=0\). This means that there does exists at least one value for \(y\) that makes the statement true, so we have proven that the original statement is true.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
does't y=0 make statement untrue?? xyzw=0, for y=0
 one year ago

nbouscalBest ResponseYou've already chosen the best response.2
Right, both the sum and the multiplication end up equaling zero because you set w=(x+z).
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
If \(y\) is 0, and \(x=(x+z)\) we have \[x+0+z+((x+z))=x+zxz=0\]and \[x\cdot0\cdot z\cdot w=0\]So that means that \[x+y+z+w=xyzw\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
Oh ... stupid me ... i never considered the negative values!!
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
Hope everyone enjoyed this little (unintuitive) problem. Good night.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
thanks!! i'll try to think broader way next time!!
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
I'll try to get a problem just for you next time then. Any preference about the subject?
 one year ago
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